Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let's consider $f, g: (0, +\infty) \rightarrow\mathbb{R}$, $f(x)=\displaystyle\frac{\sin x}{x}$, $g(x)=\displaystyle\frac{\cos x}{x}$. Find the following limits

$$\lim_{n\to\infty}f^{(n)}(x)$$ $$\lim_{n\to\infty}g^{(n)}(x)$$

where $f^{(n)}$ and $g^{(n)}$ are the $n$th derivatives of $f(x)$, respectively $g(x)$.
It's a problem I thought of last days and I didn't guess the answer by trying to look at the first derivatives of both functions. What should I do here to get the limits? Thanks.

share|improve this question
    
Do you assume this has a closed form ? –  mick Sep 29 '12 at 18:26
    
You can easily give a boundary by using taylor's theorem , but i guess you knew that already. –  mick Sep 29 '12 at 18:28
    
Maybe this is naive but if you use hypergeo form I think you can solve this. –  mick Sep 29 '12 at 18:33
    
As you take derivatives, the power of $x$ in the denominator will grow without bound. –  Graphth Sep 29 '12 at 18:35
    
Try exploiting $xy=\sin x$ and differentiate $n$ times using Leibniz's rule. –  Pedro Tamaroff Sep 29 '12 at 18:38

2 Answers 2

$f'(x) = -x^{-2}\sin x + x^{-1}\cos x$ suggests that $f^{(n)}(x)$ can be written as $P_n(x^{-1})\sin x + Q_n(x^{-1})\cos(x)$ with polynomials $P_n, Q_n\in \mathbb Z[X]$. Indeed, the derivative of $P_n(x^{-1})\sin x + Q_n(x^{-1})\cos(x)$ is $-x^{-2}P_n'(x^{-1})\sin x+P_n(x^{-1})\cos x -x^{-2}Q_n'(x^{-1})\cos(x)-Q_n(x^{-2})\sin x$, so that we are led to the recursions $$ P_{n+1}=-X^2P_n'-Q_n,\\Q_{n+1}=P_n-X^2Q_n'.$$ Letting $R_n=i^{-n}(P_n+iQ_n)\in \mathbb Z[i,X]$, we see that $$\tag1 R_{n+1}=iX^2R_n'+R_n.$$ By induction one readily shows (starting with $R_0=X$)

$$ R_n = \sum_{k=0}^ni^k\frac{n!}{(n-k)!}X^{k+1}.$$ While this allows us to write down $f^{(n)}(x)$ and $g^{(n)}(x)$ explicitly, there is no hint that $\lim_{n\to\infty}f^{(n)}(x)$ or $\lim_{n\to\infty}g^{(n)}(x)$ should exist for any $x$ (not even for multiples of $\frac\pi2$).

share|improve this answer

Consider the function $$h(x):={e^{ix}\over x}\ .$$ Computing the first few derivatives using paper and pencil one is lead to the conjecture that $$h^{(n)}(x)={1\over x}p_n\Bigl({1\over x}\Bigr)\ e^{ix}\ ,$$ where $p_n(t)=\sum_{k=0}^n c_k\ t^k$ is a polynomial of degree $\leq n$ with complex coefficients.

This conjecture can be proven by induction, and then the original claim about $f$ and $g$ is immediate.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.